What is the relationship between electric field intensity and surface charge density?
2 Answers
The relationship between **electric field intensity** (\(E\)) and **surface charge density** (\(\sigma\)) can be derived from **Gauss's Law**, which is one of Maxwell's equations in electromagnetism. This law describes how the electric field is related to the charge distribution that produces it.
### 1. **Gauss’s Law:**
In its integral form, Gauss’s law states:
\[
\oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}
\]
Where:
- \(\oint_S \mathbf{E} \cdot d\mathbf{A}\) is the electric flux through a closed surface \(S\),
- \(Q_{\text{enc}}\) is the total charge enclosed by the surface,
- \(\varepsilon_0\) is the permittivity of free space (\(8.854 \times 10^{-12} \, \text{F/m}\)).
### 2. **Surface Charge Density (\(\sigma\)):**
Surface charge density is defined as the amount of charge per unit area on a surface:
\[
\sigma = \frac{Q}{A}
\]
Where:
- \(\sigma\) is the surface charge density (in coulombs per square meter, C/m²),
- \(Q\) is the total charge on the surface,
- \(A\) is the area of the surface.
### 3. **Relationship for an Infinite Plane:**
Consider a large, uniformly charged conducting sheet or a non-conducting plane with surface charge density \(\sigma\). To find the electric field near the surface, we apply Gauss's law using a Gaussian "pillbox" surface that encloses the plane.
The pillbox will have two faces (one above and one below the charged plane) of area \(A\), with an electric field normal to the surface.
By symmetry, the electric field produced by an infinite plane of charge is uniform and perpendicular to the surface. Applying Gauss's law:
\[
\oint_S \mathbf{E} \cdot d\mathbf{A} = E \cdot A + E \cdot A = 2E \cdot A
\]
Since the charge enclosed by the Gaussian surface is \(\sigma A\), Gauss’s law becomes:
\[
2E \cdot A = \frac{\sigma A}{\varepsilon_0}
\]
Solving for \(E\):
\[
E = \frac{\sigma}{2\varepsilon_0}
\]
Thus, the **electric field intensity** just outside a uniformly charged infinite plane is directly proportional to the **surface charge density** \(\sigma\), and the relationship is:
\[
E = \frac{\sigma}{2\varepsilon_0}
\]
### 4. **Relationship for a Conducting Surface:**
For a conducting surface, the electric field just outside the surface due to a surface charge is:
\[
E = \frac{\sigma}{\varepsilon_0}
\]
This result is because, for a conductor, the field inside the conductor is zero, and all the charge resides on the surface. The electric field points perpendicular to the surface.
### Summary:
- For an **infinite plane**, the electric field intensity is \(E = \frac{\sigma}{2\varepsilon_0}\).
- For a **conducting surface**, the electric field intensity just outside is \(E = \frac{\sigma}{\varepsilon_0}\).
In both cases, the electric field is directly proportional to the surface charge density. The proportionality constant involves the permittivity of free space, \(\varepsilon_0\).
### 1. **Gauss’s Law:**
In its integral form, Gauss’s law states:
\[
\oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}
\]
Where:
- \(\oint_S \mathbf{E} \cdot d\mathbf{A}\) is the electric flux through a closed surface \(S\),
- \(Q_{\text{enc}}\) is the total charge enclosed by the surface,
- \(\varepsilon_0\) is the permittivity of free space (\(8.854 \times 10^{-12} \, \text{F/m}\)).
### 2. **Surface Charge Density (\(\sigma\)):**
Surface charge density is defined as the amount of charge per unit area on a surface:
\[
\sigma = \frac{Q}{A}
\]
Where:
- \(\sigma\) is the surface charge density (in coulombs per square meter, C/m²),
- \(Q\) is the total charge on the surface,
- \(A\) is the area of the surface.
### 3. **Relationship for an Infinite Plane:**
Consider a large, uniformly charged conducting sheet or a non-conducting plane with surface charge density \(\sigma\). To find the electric field near the surface, we apply Gauss's law using a Gaussian "pillbox" surface that encloses the plane.
The pillbox will have two faces (one above and one below the charged plane) of area \(A\), with an electric field normal to the surface.
By symmetry, the electric field produced by an infinite plane of charge is uniform and perpendicular to the surface. Applying Gauss's law:
\[
\oint_S \mathbf{E} \cdot d\mathbf{A} = E \cdot A + E \cdot A = 2E \cdot A
\]
Since the charge enclosed by the Gaussian surface is \(\sigma A\), Gauss’s law becomes:
\[
2E \cdot A = \frac{\sigma A}{\varepsilon_0}
\]
Solving for \(E\):
\[
E = \frac{\sigma}{2\varepsilon_0}
\]
Thus, the **electric field intensity** just outside a uniformly charged infinite plane is directly proportional to the **surface charge density** \(\sigma\), and the relationship is:
\[
E = \frac{\sigma}{2\varepsilon_0}
\]
### 4. **Relationship for a Conducting Surface:**
For a conducting surface, the electric field just outside the surface due to a surface charge is:
\[
E = \frac{\sigma}{\varepsilon_0}
\]
This result is because, for a conductor, the field inside the conductor is zero, and all the charge resides on the surface. The electric field points perpendicular to the surface.
### Summary:
- For an **infinite plane**, the electric field intensity is \(E = \frac{\sigma}{2\varepsilon_0}\).
- For a **conducting surface**, the electric field intensity just outside is \(E = \frac{\sigma}{\varepsilon_0}\).
In both cases, the electric field is directly proportional to the surface charge density. The proportionality constant involves the permittivity of free space, \(\varepsilon_0\).
The relationship between electric field intensity (**E**) and surface charge density (**σ**) is fundamental in electrostatics, especially when discussing conductors and surfaces.
### Electric Field Intensity (**E**):
Electric field intensity is a vector quantity that represents the force per unit charge exerted on a small positive test charge placed in the field. It is expressed as:
\[
E = \frac{F}{q}
\]
Where:
- **E** is the electric field intensity (in volts per meter, V/m),
- **F** is the force experienced by the charge (in newtons, N),
- **q** is the magnitude of the charge (in coulombs, C).
### Surface Charge Density (**σ**):
Surface charge density is a measure of how much electric charge is accumulated per unit area on a surface. It is defined as:
\[
\sigma = \frac{Q}{A}
\]
Where:
- **σ** is the surface charge density (in coulombs per square meter, C/m²),
- **Q** is the charge distributed over the surface (in coulombs, C),
- **A** is the area of the surface (in square meters, m²).
### Relationship Between Electric Field Intensity and Surface Charge Density:
When dealing with an infinite plane of charge or a large conducting surface, the relationship between the electric field and the surface charge density can be derived from **Gauss’s Law**, which states:
\[
\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}
\]
Where:
- **\( \oint \vec{E} \cdot d\vec{A} \)** is the electric flux through a Gaussian surface,
- **\( Q_{\text{enclosed}} \)** is the charge enclosed by the surface,
- **\( \varepsilon_0 \)** is the permittivity of free space (\( 8.854 \times 10^{-12} \) F/m).
For a flat conductor or an infinite charged plane, the electric field just outside the surface is related to the surface charge density by the following equation:
\[
E = \frac{\sigma}{\varepsilon_0}
\]
Where:
- **E** is the electric field just outside the surface (in V/m),
- **σ** is the surface charge density (in C/m²),
- **\( \varepsilon_0 \)** is the permittivity of free space.
### Key Insights:
1. **Direct Proportionality**: The electric field intensity is directly proportional to the surface charge density. If the charge density increases, the electric field intensity increases as well.
2. **Uniform Electric Field**: For a large, uniformly charged surface, the electric field is uniform and perpendicular to the surface.
### Example:
Consider a large, flat conducting plate with a surface charge density of \( 1 \times 10^{-6} \, \text{C/m}^2 \). The electric field just outside the surface would be:
\[
E = \frac{1 \times 10^{-6}}{8.854 \times 10^{-12}} \, \text{V/m} = 1.13 \times 10^5 \, \text{V/m}
\]
Thus, the electric field intensity at the surface of the plate is \( 1.13 \times 10^5 \, \text{V/m} \).
In summary, the electric field intensity near a surface is proportional to the surface charge density, and the relationship is governed by Gauss's law.
### Electric Field Intensity (**E**):
Electric field intensity is a vector quantity that represents the force per unit charge exerted on a small positive test charge placed in the field. It is expressed as:
\[
E = \frac{F}{q}
\]
Where:
- **E** is the electric field intensity (in volts per meter, V/m),
- **F** is the force experienced by the charge (in newtons, N),
- **q** is the magnitude of the charge (in coulombs, C).
### Surface Charge Density (**σ**):
Surface charge density is a measure of how much electric charge is accumulated per unit area on a surface. It is defined as:
\[
\sigma = \frac{Q}{A}
\]
Where:
- **σ** is the surface charge density (in coulombs per square meter, C/m²),
- **Q** is the charge distributed over the surface (in coulombs, C),
- **A** is the area of the surface (in square meters, m²).
### Relationship Between Electric Field Intensity and Surface Charge Density:
When dealing with an infinite plane of charge or a large conducting surface, the relationship between the electric field and the surface charge density can be derived from **Gauss’s Law**, which states:
\[
\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}
\]
Where:
- **\( \oint \vec{E} \cdot d\vec{A} \)** is the electric flux through a Gaussian surface,
- **\( Q_{\text{enclosed}} \)** is the charge enclosed by the surface,
- **\( \varepsilon_0 \)** is the permittivity of free space (\( 8.854 \times 10^{-12} \) F/m).
For a flat conductor or an infinite charged plane, the electric field just outside the surface is related to the surface charge density by the following equation:
\[
E = \frac{\sigma}{\varepsilon_0}
\]
Where:
- **E** is the electric field just outside the surface (in V/m),
- **σ** is the surface charge density (in C/m²),
- **\( \varepsilon_0 \)** is the permittivity of free space.
### Key Insights:
1. **Direct Proportionality**: The electric field intensity is directly proportional to the surface charge density. If the charge density increases, the electric field intensity increases as well.
2. **Uniform Electric Field**: For a large, uniformly charged surface, the electric field is uniform and perpendicular to the surface.
### Example:
Consider a large, flat conducting plate with a surface charge density of \( 1 \times 10^{-6} \, \text{C/m}^2 \). The electric field just outside the surface would be:
\[
E = \frac{1 \times 10^{-6}}{8.854 \times 10^{-12}} \, \text{V/m} = 1.13 \times 10^5 \, \text{V/m}
\]
Thus, the electric field intensity at the surface of the plate is \( 1.13 \times 10^5 \, \text{V/m} \).
In summary, the electric field intensity near a surface is proportional to the surface charge density, and the relationship is governed by Gauss's law.